Question

Consider aqueous solutions of the following coordination compounds: \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{I}_{3}, \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{4}, \mathrm{Na}_{2} \mathrm{Pt}_{6}\), and \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{3} .\) If aqueous \(\mathrm{AgNO}_{3}\) is added to separate beakers containing solutions of each coordination compound, how many moles of AgI will precipitate per mole of transition metal present? Assume that each transition metal ion forms an octahedral complex.

Short answer

The number of moles of AgI formed per mole of transition metal for each coordination compound: 1. \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{I}_{3}\): 3 moles of AgI per mole of Co 2. \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{4}\): 4 moles of AgI per mole of Pt 3. \(\mathrm{Na}_{2} \mathrm{Pt}_{6}\): 0 moles of AgI (no iodide ions present) 4. \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{3}\): 3 moles of AgI per mole of Cr

Step-by-step solution

Identify the transition metals and the number of iodide ions in each coordination compound

For each coordination compound, identify the transition metal (TM) and the number of iodide ions (I) bound to it. 1. \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{I}_{3}\): TM = Co, I = 3 2. \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{4}\): TM = Pt, I = 4 3. \(\mathrm{Na}_{2} \mathrm{Pt}_{6}\): Since there is no iodide ion present in this compound, it will not form any AgI precipitate. 4. \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{3}\): TM = Cr, I = 3 #Step 2: Determine the number of moles of AgI formed per mole of transition metal#

Calculate the moles of AgI formed per mole of transition metal

For each coordination compound that has iodide ions, determine the moles of AgI formed per mole of transition metal. 1. \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{I}_{3}\): 3 moles of I per mole of Co. Since the stoichiometry between Ag+ and I- is 1:1, there will be 3 moles of AgI formed per mole of Co. 2. \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{4}\): 4 moles of I per mole of Pt. Since the stoichiometry between Ag+ and I- is 1:1, there will be 4 moles of AgI formed per mole of Pt. 3. \(\mathrm{Na}_{2} \mathrm{Pt}_{6}\): No AgI formed as there are no iodide ions present. 4. \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{3}\): 3 moles of I per mole of Cr. Since the stoichiometry between Ag+ and I- is 1:1, there will be 3 moles of AgI formed per mole of Cr. #Step 3: Present the final results#

Provide the final answer

The number of moles of AgI formed per mole of transition metal for each coordination compound: 1. \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{I}_{3}\): 3 moles of AgI per mole of Co 2. \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{4}\): 4 moles of AgI per mole of Pt 3. \(\mathrm{Na}_{2} \mathrm{Pt}_{6}\): 0 moles of AgI (no iodide ions present) 4. \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{3}\): 3 moles of AgI per mole of Cr And that's the solution for this exercise.

Key concepts

Transition Metal Complexes

Transition metal complexes are chemical compounds formed between transition metals and ligands where ligands are atoms, ions, or molecules that can donate a pair of electrons to the metal. These complexes exhibit a variety of structures, but one common architecture is the octahedral formation, where six ligands symmetrically surround a transition metal ion.

In the exercise, compounds like \(\mathrm{Co}(\mathrm{NH}_{3})_{6} \mathrm{I}_{3}\), \(\mathrm{Pt}(\mathrm{NH}_{3})_{4} \mathrm{I}_{4}\), and \(\mathrm{Cr}(\mathrm{NH}_{3})_{4} \mathrm{I}_{3}\) illustrate this octahedral complex. The ammonium (\(\mathrm{NH}_{3}\)) serves as a neutral ligand, while iodide ions (I-) act as anionic ligands that can potentially dissociate in a reaction. The process of forming these complexes involves coordinate covalent bonds where ligands share electron pairs with the metal center.

Understanding these complexes' structures aids in predicting their chemical behavior, such as reactivity and interaction with other compounds in solution. This is essential in the exercise when considering the precipitation reactions that will occur upon the addition of \(\mathrm{AgNO}_{3}\) to the solutions of each complex.

Stoichiometry

Stoichiometry refers to the calculation of reactants and products in chemical reactions. It is a quantitative measure of the relative amounts of substances involved in a reaction. In the context of the exercise, stoichiometry is used to determine the number of moles of silver iodide (AgI) that would precipitate per mole of transition metal in the solution when \(\mathrm{AgNO}_{3}\) is added.

The key to solving stoichiometric problems is to understand the molar ratios of the reactants and products. Here, the stoichiometry of the reaction between the silver cation (\(\mathrm{Ag}^+\)) and iodide ion (\(\mathrm{I}^-\)) is 1:1, meaning one silver ion will react with one iodide ion to form one molecule of AgI. With this knowledge, you can easily link the number of iodide ions in each complex to the moles of AgI's precipitation.

To improve upon common textbook solutions, we should demonstrate the importance of balancing chemical equations and maintaining the conservation of mass in our calculations. This not only helps in getting the correct answer but also fosters a deeper understanding of the underlying principles of chemistry.

Precipitation Reactions

Precipitation reactions are a type of chemical reaction where an insoluble solid, known as a precipitate, is formed from the reaction of ions in solution. The formation of a precipitate indicates that a chemical change has taken place. The insoluble solid emerges from the solution, can be collected by filtration, and its mass can be determined.

In the textbook exercise, we examine how aqueous \(\mathrm{AgNO}_{3}\) reacts with coordination compounds containing iodide ions. When \(\mathrm{Ag}^+\) ions are introduced to an aqueous solution containing iodide ions, they will spontaneously react to form silver iodide (AgI), which is a yellow, insoluble solid.

Predicting a Precipitate

One can predict whether a precipitation reaction will occur by consulting the solubility rules. Silver iodide is notably insoluble in water, making it an ideal candidate for such reactions.

Relevance to the Exercise

In our case, the \(\mathrm{Ag}^+\) ions from \(\mathrm{AgNO}_{3}\) will pair with the dissociated iodide ions from the coordination compounds to form AgI precipitates. By understanding the concept of precipitation and the solubility of various compounds, students are equipped to perform calculations and predict the outcomes of reactions which is crucial for successfully executing exercises such as the one provided.